/**
 * Licensed to the Apache Software Foundation (ASF) under one or more
 * contributor license agreements.  See the NOTICE file distributed with
 * this work for additional information regarding copyright ownership.
 * The ASF licenses this file to You under the Apache License, Version 2.0
 * (the "License"); you may not use this file except in compliance with
 * the License.  You may obtain a copy of the License at
 *
 *     http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */

package org.apache.lucene.util; // from org.apache.solr.util rev 555343

/**  A variety of high efficiency bit twiddling routines.
 * @lucene.internal
 */
public final class BitUtil {

    private BitUtil() {
    } // no instance

    /** Returns the number of bits set in the long */
    public static int pop(long x) {
        /* Hacker's Delight 32 bit pop function:
         * http://www.hackersdelight.org/HDcode/newCode/pop_arrayHS.cc
         *
        int pop(unsigned x) {
           x = x - ((x >> 1) & 0x55555555);
           x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
           x = (x + (x >> 4)) & 0x0F0F0F0F;
           x = x + (x >> 8);
           x = x + (x >> 16);
           return x & 0x0000003F;
          }
        ***/

        // 64 bit java version of the C function from above
        x = x - ((x >>> 1) & 0x5555555555555555L);
        x = (x & 0x3333333333333333L) + ((x >>> 2) & 0x3333333333333333L);
        x = (x + (x >>> 4)) & 0x0F0F0F0F0F0F0F0FL;
        x = x + (x >>> 8);
        x = x + (x >>> 16);
        x = x + (x >>> 32);
        return ((int) x) & 0x7F;
    }

    /*** Returns the number of set bits in an array of longs. */
    public static long pop_array(long A[], int wordOffset, int numWords) {
        /*
        * Robert Harley and David Seal's bit counting algorithm, as documented
        * in the revisions of Hacker's Delight
        * http://www.hackersdelight.org/revisions.pdf
        * http://www.hackersdelight.org/HDcode/newCode/pop_arrayHS.cc
        *
        * This function was adapted to Java, and extended to use 64 bit words.
        * if only we had access to wider registers like SSE from java...
        *
        * This function can be transformed to compute the popcount of other functions
        * on bitsets via something like this:
        * sed 's/A\[\([^]]*\)\]/\(A[\1] \& B[\1]\)/g'
        *
        */
        int n = wordOffset + numWords;
        long tot = 0, tot8 = 0;
        long ones = 0, twos = 0, fours = 0;

        int i;
        for (i = wordOffset; i <= n - 8; i += 8) {
            /***  C macro from Hacker's Delight
             #define CSA(h,l, a,b,c) \
             {unsigned u = a ^ b; unsigned v = c; \
             h = (a & b) | (u & v); l = u ^ v;}
             ***/

            long twosA, twosB, foursA, foursB, eights;

            // CSA(twosA, ones, ones, A[i], A[i+1])
            {
                long b = A[i], c = A[i + 1];
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            // CSA(twosB, ones, ones, A[i+2], A[i+3])
            {
                long b = A[i + 2], c = A[i + 3];
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            //CSA(foursA, twos, twos, twosA, twosB)
            {
                long u = twos ^ twosA;
                foursA = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }
            //CSA(twosA, ones, ones, A[i+4], A[i+5])
            {
                long b = A[i + 4], c = A[i + 5];
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            // CSA(twosB, ones, ones, A[i+6], A[i+7])
            {
                long b = A[i + 6], c = A[i + 7];
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            //CSA(foursB, twos, twos, twosA, twosB)
            {
                long u = twos ^ twosA;
                foursB = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }

            //CSA(eights, fours, fours, foursA, foursB)
            {
                long u = fours ^ foursA;
                eights = (fours & foursA) | (u & foursB);
                fours = u ^ foursB;
            }
            tot8 += pop(eights);
        }

        // handle trailing words in a binary-search manner...
        // derived from the loop above by setting specific elements to 0.
        // the original method in Hackers Delight used a simple for loop:
        //   for (i = i; i < n; i++)      // Add in the last elements
        //  tot = tot + pop(A[i]);

        if (i <= n - 4) {
            long twosA, twosB, foursA, eights;
            {
                long b = A[i], c = A[i + 1];
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            {
                long b = A[i + 2], c = A[i + 3];
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            {
                long u = twos ^ twosA;
                foursA = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }
            eights = fours & foursA;
            fours = fours ^ foursA;

            tot8 += pop(eights);
            i += 4;
        }

        if (i <= n - 2) {
            long b = A[i], c = A[i + 1];
            long u = ones ^ b;
            long twosA = (ones & b) | (u & c);
            ones = u ^ c;

            long foursA = twos & twosA;
            twos = twos ^ twosA;

            long eights = fours & foursA;
            fours = fours ^ foursA;

            tot8 += pop(eights);
            i += 2;
        }

        if (i < n) {
            tot += pop(A[i]);
        }

        tot += (pop(fours) << 2) + (pop(twos) << 1) + pop(ones) + (tot8 << 3);

        return tot;
    }

    /** Returns the popcount or cardinality of the two sets after an intersection.
     * Neither array is modified.
     */
    public static long pop_intersect(long A[], long B[], int wordOffset, int numWords) {
        // generated from pop_array via sed 's/A\[\([^]]*\)\]/\(A[\1] \& B[\1]\)/g'
        int n = wordOffset + numWords;
        long tot = 0, tot8 = 0;
        long ones = 0, twos = 0, fours = 0;

        int i;
        for (i = wordOffset; i <= n - 8; i += 8) {
            long twosA, twosB, foursA, foursB, eights;

            // CSA(twosA, ones, ones, (A[i] & B[i]), (A[i+1] & B[i+1]))
            {
                long b = (A[i] & B[i]), c = (A[i + 1] & B[i + 1]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            // CSA(twosB, ones, ones, (A[i+2] & B[i+2]), (A[i+3] & B[i+3]))
            {
                long b = (A[i + 2] & B[i + 2]), c = (A[i + 3] & B[i + 3]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            //CSA(foursA, twos, twos, twosA, twosB)
            {
                long u = twos ^ twosA;
                foursA = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }
            //CSA(twosA, ones, ones, (A[i+4] & B[i+4]), (A[i+5] & B[i+5]))
            {
                long b = (A[i + 4] & B[i + 4]), c = (A[i + 5] & B[i + 5]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            // CSA(twosB, ones, ones, (A[i+6] & B[i+6]), (A[i+7] & B[i+7]))
            {
                long b = (A[i + 6] & B[i + 6]), c = (A[i + 7] & B[i + 7]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            //CSA(foursB, twos, twos, twosA, twosB)
            {
                long u = twos ^ twosA;
                foursB = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }

            //CSA(eights, fours, fours, foursA, foursB)
            {
                long u = fours ^ foursA;
                eights = (fours & foursA) | (u & foursB);
                fours = u ^ foursB;
            }
            tot8 += pop(eights);
        }

        if (i <= n - 4) {
            long twosA, twosB, foursA, eights;
            {
                long b = (A[i] & B[i]), c = (A[i + 1] & B[i + 1]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            {
                long b = (A[i + 2] & B[i + 2]), c = (A[i + 3] & B[i + 3]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            {
                long u = twos ^ twosA;
                foursA = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }
            eights = fours & foursA;
            fours = fours ^ foursA;

            tot8 += pop(eights);
            i += 4;
        }

        if (i <= n - 2) {
            long b = (A[i] & B[i]), c = (A[i + 1] & B[i + 1]);
            long u = ones ^ b;
            long twosA = (ones & b) | (u & c);
            ones = u ^ c;

            long foursA = twos & twosA;
            twos = twos ^ twosA;

            long eights = fours & foursA;
            fours = fours ^ foursA;

            tot8 += pop(eights);
            i += 2;
        }

        if (i < n) {
            tot += pop((A[i] & B[i]));
        }

        tot += (pop(fours) << 2) + (pop(twos) << 1) + pop(ones) + (tot8 << 3);

        return tot;
    }

    /** Returns the popcount or cardinality of the union of two sets.
    * Neither array is modified.
    */
    public static long pop_union(long A[], long B[], int wordOffset, int numWords) {
        // generated from pop_array via sed 's/A\[\([^]]*\)\]/\(A[\1] \| B[\1]\)/g'
        int n = wordOffset + numWords;
        long tot = 0, tot8 = 0;
        long ones = 0, twos = 0, fours = 0;

        int i;
        for (i = wordOffset; i <= n - 8; i += 8) {
            /***  C macro from Hacker's Delight
             #define CSA(h,l, a,b,c) \
             {unsigned u = a ^ b; unsigned v = c; \
             h = (a & b) | (u & v); l = u ^ v;}
             ***/

            long twosA, twosB, foursA, foursB, eights;

            // CSA(twosA, ones, ones, (A[i] | B[i]), (A[i+1] | B[i+1]))
            {
                long b = (A[i] | B[i]), c = (A[i + 1] | B[i + 1]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            // CSA(twosB, ones, ones, (A[i+2] | B[i+2]), (A[i+3] | B[i+3]))
            {
                long b = (A[i + 2] | B[i + 2]), c = (A[i + 3] | B[i + 3]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            //CSA(foursA, twos, twos, twosA, twosB)
            {
                long u = twos ^ twosA;
                foursA = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }
            //CSA(twosA, ones, ones, (A[i+4] | B[i+4]), (A[i+5] | B[i+5]))
            {
                long b = (A[i + 4] | B[i + 4]), c = (A[i + 5] | B[i + 5]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            // CSA(twosB, ones, ones, (A[i+6] | B[i+6]), (A[i+7] | B[i+7]))
            {
                long b = (A[i + 6] | B[i + 6]), c = (A[i + 7] | B[i + 7]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            //CSA(foursB, twos, twos, twosA, twosB)
            {
                long u = twos ^ twosA;
                foursB = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }

            //CSA(eights, fours, fours, foursA, foursB)
            {
                long u = fours ^ foursA;
                eights = (fours & foursA) | (u & foursB);
                fours = u ^ foursB;
            }
            tot8 += pop(eights);
        }

        if (i <= n - 4) {
            long twosA, twosB, foursA, eights;
            {
                long b = (A[i] | B[i]), c = (A[i + 1] | B[i + 1]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            {
                long b = (A[i + 2] | B[i + 2]), c = (A[i + 3] | B[i + 3]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            {
                long u = twos ^ twosA;
                foursA = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }
            eights = fours & foursA;
            fours = fours ^ foursA;

            tot8 += pop(eights);
            i += 4;
        }

        if (i <= n - 2) {
            long b = (A[i] | B[i]), c = (A[i + 1] | B[i + 1]);
            long u = ones ^ b;
            long twosA = (ones & b) | (u & c);
            ones = u ^ c;

            long foursA = twos & twosA;
            twos = twos ^ twosA;

            long eights = fours & foursA;
            fours = fours ^ foursA;

            tot8 += pop(eights);
            i += 2;
        }

        if (i < n) {
            tot += pop((A[i] | B[i]));
        }

        tot += (pop(fours) << 2) + (pop(twos) << 1) + pop(ones) + (tot8 << 3);

        return tot;
    }

    /** Returns the popcount or cardinality of A & ~B
     * Neither array is modified.
     */
    public static long pop_andnot(long A[], long B[], int wordOffset, int numWords) {
        // generated from pop_array via sed 's/A\[\([^]]*\)\]/\(A[\1] \& ~B[\1]\)/g'
        int n = wordOffset + numWords;
        long tot = 0, tot8 = 0;
        long ones = 0, twos = 0, fours = 0;

        int i;
        for (i = wordOffset; i <= n - 8; i += 8) {
            /***  C macro from Hacker's Delight
             #define CSA(h,l, a,b,c) \
             {unsigned u = a ^ b; unsigned v = c; \
             h = (a & b) | (u & v); l = u ^ v;}
             ***/

            long twosA, twosB, foursA, foursB, eights;

            // CSA(twosA, ones, ones, (A[i] & ~B[i]), (A[i+1] & ~B[i+1]))
            {
                long b = (A[i] & ~B[i]), c = (A[i + 1] & ~B[i + 1]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            // CSA(twosB, ones, ones, (A[i+2] & ~B[i+2]), (A[i+3] & ~B[i+3]))
            {
                long b = (A[i + 2] & ~B[i + 2]), c = (A[i + 3] & ~B[i + 3]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            //CSA(foursA, twos, twos, twosA, twosB)
            {
                long u = twos ^ twosA;
                foursA = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }
            //CSA(twosA, ones, ones, (A[i+4] & ~B[i+4]), (A[i+5] & ~B[i+5]))
            {
                long b = (A[i + 4] & ~B[i + 4]), c = (A[i + 5] & ~B[i + 5]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            // CSA(twosB, ones, ones, (A[i+6] & ~B[i+6]), (A[i+7] & ~B[i+7]))
            {
                long b = (A[i + 6] & ~B[i + 6]), c = (A[i + 7] & ~B[i + 7]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            //CSA(foursB, twos, twos, twosA, twosB)
            {
                long u = twos ^ twosA;
                foursB = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }

            //CSA(eights, fours, fours, foursA, foursB)
            {
                long u = fours ^ foursA;
                eights = (fours & foursA) | (u & foursB);
                fours = u ^ foursB;
            }
            tot8 += pop(eights);
        }

        if (i <= n - 4) {
            long twosA, twosB, foursA, eights;
            {
                long b = (A[i] & ~B[i]), c = (A[i + 1] & ~B[i + 1]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            {
                long b = (A[i + 2] & ~B[i + 2]), c = (A[i + 3] & ~B[i + 3]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            {
                long u = twos ^ twosA;
                foursA = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }
            eights = fours & foursA;
            fours = fours ^ foursA;

            tot8 += pop(eights);
            i += 4;
        }

        if (i <= n - 2) {
            long b = (A[i] & ~B[i]), c = (A[i + 1] & ~B[i + 1]);
            long u = ones ^ b;
            long twosA = (ones & b) | (u & c);
            ones = u ^ c;

            long foursA = twos & twosA;
            twos = twos ^ twosA;

            long eights = fours & foursA;
            fours = fours ^ foursA;

            tot8 += pop(eights);
            i += 2;
        }

        if (i < n) {
            tot += pop((A[i] & ~B[i]));
        }

        tot += (pop(fours) << 2) + (pop(twos) << 1) + pop(ones) + (tot8 << 3);

        return tot;
    }

    public static long pop_xor(long A[], long B[], int wordOffset, int numWords) {
        int n = wordOffset + numWords;
        long tot = 0, tot8 = 0;
        long ones = 0, twos = 0, fours = 0;

        int i;
        for (i = wordOffset; i <= n - 8; i += 8) {
            /***  C macro from Hacker's Delight
             #define CSA(h,l, a,b,c) \
             {unsigned u = a ^ b; unsigned v = c; \
             h = (a & b) | (u & v); l = u ^ v;}
             ***/

            long twosA, twosB, foursA, foursB, eights;

            // CSA(twosA, ones, ones, (A[i] ^ B[i]), (A[i+1] ^ B[i+1]))
            {
                long b = (A[i] ^ B[i]), c = (A[i + 1] ^ B[i + 1]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            // CSA(twosB, ones, ones, (A[i+2] ^ B[i+2]), (A[i+3] ^ B[i+3]))
            {
                long b = (A[i + 2] ^ B[i + 2]), c = (A[i + 3] ^ B[i + 3]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            //CSA(foursA, twos, twos, twosA, twosB)
            {
                long u = twos ^ twosA;
                foursA = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }
            //CSA(twosA, ones, ones, (A[i+4] ^ B[i+4]), (A[i+5] ^ B[i+5]))
            {
                long b = (A[i + 4] ^ B[i + 4]), c = (A[i + 5] ^ B[i + 5]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            // CSA(twosB, ones, ones, (A[i+6] ^ B[i+6]), (A[i+7] ^ B[i+7]))
            {
                long b = (A[i + 6] ^ B[i + 6]), c = (A[i + 7] ^ B[i + 7]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            //CSA(foursB, twos, twos, twosA, twosB)
            {
                long u = twos ^ twosA;
                foursB = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }

            //CSA(eights, fours, fours, foursA, foursB)
            {
                long u = fours ^ foursA;
                eights = (fours & foursA) | (u & foursB);
                fours = u ^ foursB;
            }
            tot8 += pop(eights);
        }

        if (i <= n - 4) {
            long twosA, twosB, foursA, eights;
            {
                long b = (A[i] ^ B[i]), c = (A[i + 1] ^ B[i + 1]);
                long u = ones ^ b;
                twosA = (ones & b) | (u & c);
                ones = u ^ c;
            }
            {
                long b = (A[i + 2] ^ B[i + 2]), c = (A[i + 3] ^ B[i + 3]);
                long u = ones ^ b;
                twosB = (ones & b) | (u & c);
                ones = u ^ c;
            }
            {
                long u = twos ^ twosA;
                foursA = (twos & twosA) | (u & twosB);
                twos = u ^ twosB;
            }
            eights = fours & foursA;
            fours = fours ^ foursA;

            tot8 += pop(eights);
            i += 4;
        }

        if (i <= n - 2) {
            long b = (A[i] ^ B[i]), c = (A[i + 1] ^ B[i + 1]);
            long u = ones ^ b;
            long twosA = (ones & b) | (u & c);
            ones = u ^ c;

            long foursA = twos & twosA;
            twos = twos ^ twosA;

            long eights = fours & foursA;
            fours = fours ^ foursA;

            tot8 += pop(eights);
            i += 2;
        }

        if (i < n) {
            tot += pop((A[i] ^ B[i]));
        }

        tot += (pop(fours) << 2) + (pop(twos) << 1) + pop(ones) + (tot8 << 3);

        return tot;
    }

    /* python code to generate ntzTable
    def ntz(val):
    if val==0: return 8
    i=0
    while (val&0x01)==0:
      i = i+1
      val >>= 1
    return i
    print ','.join([ str(ntz(i)) for i in range(256) ])
    ***/
    /** table of number of trailing zeros in a byte */
    public static final byte[] ntzTable = { 8, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 6, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 7, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 6, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0 };

    /** Returns number of trailing zeros in a 64 bit long value. */
    public static int ntz(long val) {
        // A full binary search to determine the low byte was slower than
        // a linear search for nextSetBit().  This is most likely because
        // the implementation of nextSetBit() shifts bits to the right, increasing
        // the probability that the first non-zero byte is in the rhs.
        //
        // This implementation does a single binary search at the top level only
        // so that all other bit shifting can be done on ints instead of longs to
        // remain friendly to 32 bit architectures.  In addition, the case of a
        // non-zero first byte is checked for first because it is the most common
        // in dense bit arrays.

        int lower = (int) val;
        int lowByte = lower & 0xff;
        if (lowByte != 0)
            return ntzTable[lowByte];

        if (lower != 0) {
            lowByte = (lower >>> 8) & 0xff;
            if (lowByte != 0)
                return ntzTable[lowByte] + 8;
            lowByte = (lower >>> 16) & 0xff;
            if (lowByte != 0)
                return ntzTable[lowByte] + 16;
            // no need to mask off low byte for the last byte in the 32 bit word
            // no need to check for zero on the last byte either.
            return ntzTable[lower >>> 24] + 24;
        } else {
            // grab upper 32 bits
            int upper = (int) (val >> 32);
            lowByte = upper & 0xff;
            if (lowByte != 0)
                return ntzTable[lowByte] + 32;
            lowByte = (upper >>> 8) & 0xff;
            if (lowByte != 0)
                return ntzTable[lowByte] + 40;
            lowByte = (upper >>> 16) & 0xff;
            if (lowByte != 0)
                return ntzTable[lowByte] + 48;
            // no need to mask off low byte for the last byte in the 32 bit word
            // no need to check for zero on the last byte either.
            return ntzTable[upper >>> 24] + 56;
        }
    }

    /** Returns number of trailing zeros in a 32 bit int value. */
    public static int ntz(int val) {
        // This implementation does a single binary search at the top level only.
        // In addition, the case of a non-zero first byte is checked for first
        // because it is the most common in dense bit arrays.

        int lowByte = val & 0xff;
        if (lowByte != 0)
            return ntzTable[lowByte];
        lowByte = (val >>> 8) & 0xff;
        if (lowByte != 0)
            return ntzTable[lowByte] + 8;
        lowByte = (val >>> 16) & 0xff;
        if (lowByte != 0)
            return ntzTable[lowByte] + 16;
        // no need to mask off low byte for the last byte.
        // no need to check for zero on the last byte either.
        return ntzTable[val >>> 24] + 24;
    }

    /** returns 0 based index of first set bit
     * (only works for x!=0)
     * <br/> This is an alternate implementation of ntz()
     */
    public static int ntz2(long x) {
        int n = 0;
        int y = (int) x;
        if (y == 0) {
            n += 32;
            y = (int) (x >>> 32);
        } // the only 64 bit shift necessary
        if ((y & 0x0000FFFF) == 0) {
            n += 16;
            y >>>= 16;
        }
        if ((y & 0x000000FF) == 0) {
            n += 8;
            y >>>= 8;
        }
        return (ntzTable[y & 0xff]) + n;
    }

    /** returns 0 based index of first set bit
     * <br/> This is an alternate implementation of ntz()
     */
    public static int ntz3(long x) {
        // another implementation taken from Hackers Delight, extended to 64 bits
        // and converted to Java.
        // Many 32 bit ntz algorithms are at http://www.hackersdelight.org/HDcode/ntz.cc
        int n = 1;

        // do the first step as a long, all others as ints.
        int y = (int) x;
        if (y == 0) {
            n += 32;
            y = (int) (x >>> 32);
        }
        if ((y & 0x0000FFFF) == 0) {
            n += 16;
            y >>>= 16;
        }
        if ((y & 0x000000FF) == 0) {
            n += 8;
            y >>>= 8;
        }
        if ((y & 0x0000000F) == 0) {
            n += 4;
            y >>>= 4;
        }
        if ((y & 0x00000003) == 0) {
            n += 2;
            y >>>= 2;
        }
        return n - (y & 1);
    }

    /** table of number of leading zeros in a byte */
    public static final byte[] nlzTable = { 8, 7, 6, 6, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 };

    /** Returns the number of leading zero bits.
     */
    public static int nlz(long x) {
        int n = 0;
        // do the first step as a long
        int y = (int) (x >>> 32);
        if (y == 0) {
            n += 32;
            y = (int) (x);
        }
        if ((y & 0xFFFF0000) == 0) {
            n += 16;
            y <<= 16;
        }
        if ((y & 0xFF000000) == 0) {
            n += 8;
            y <<= 8;
        }
        return n + nlzTable[y >>> 24];
        /* implementation without table:
          if ((y & 0xF0000000) == 0) { n+=4; y<<=4; }
          if ((y & 0xC0000000) == 0) { n+=2; y<<=2; }
          if ((y & 0x80000000) == 0) { n+=1; y<<=1; }
          if ((y & 0x80000000) == 0) { n+=1;}
          return n;
         */
    }

    /** returns true if v is a power of two or zero*/
    public static boolean isPowerOfTwo(int v) {
        return ((v & (v - 1)) == 0);
    }

    /** returns true if v is a power of two or zero*/
    public static boolean isPowerOfTwo(long v) {
        return ((v & (v - 1)) == 0);
    }

    /** returns the next highest power of two, or the current value if it's already a power of two or zero*/
    public static int nextHighestPowerOfTwo(int v) {
        v--;
        v |= v >> 1;
        v |= v >> 2;
        v |= v >> 4;
        v |= v >> 8;
        v |= v >> 16;
        v++;
        return v;
    }

    /** returns the next highest power of two, or the current value if it's already a power of two or zero*/
    public static long nextHighestPowerOfTwo(long v) {
        v--;
        v |= v >> 1;
        v |= v >> 2;
        v |= v >> 4;
        v |= v >> 8;
        v |= v >> 16;
        v |= v >> 32;
        v++;
        return v;
    }

}
